Conservation of energy with a mass and pulley system.

  • Thread starter sdoi
  • Start date
  • #1
37
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Homework Statement


Two boxes are attached to opposite ends of a rope passing over a frictionless pulley as shown below. The mass of Box A is 15kg and the mass of box B is 12kg. The system is originally at rest with the bottom of box A at a height of o.85m above the floor. When the system is released, the boxes will move. Use conservation of energy to determine the speed with which Box A will contact the floor.


Homework Equations



Eg=mgΔh
Ek= 1/2 mv^2/2
ƩFy=may

The Attempt at a Solution


I started off by drawing free body diagrams of each mass, one at rest, and one in motion.

For mass A:
at rest,
ƩFy=0
Ft+Eg=0
Ft= Eg
= mgΔh
=(15kg)(9.8m/s^2)(0.85m)
Ft=125N

in motion,
ƩFy= may
Fg(A)-Ft= m(A)ay

For mass B:
at rest,
ƩFy=0
Fn-mg=0
Fn=mg
=(12kg)(9.8m/s^2)
Fn=117.6N

in motion,
ƩFy=may
Ft-Fg(B)= m(B)ay

I'm not sure if my original statement of Ft=Eg is accurate... and from this point on I don't know where to go.
 

Answers and Replies

  • #2
993
13
Why not find the initial and final energies and equate them?
 
  • #3
37
0
As in, Etotal= Eg, Etotal'=Ek ?
 
  • #4
993
13
System is originally at rest. Hence initial energy = ...
Final energy = ...
 

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